Altitudes and Orthocenters

In this assignment, I worked with Lauren Wright in exploring the altitudes and orthocenter of an acute triangle. What began as a simple "ask your office neighbor for some help" on her part, turned into an obsessive need to find an answer to a geometric proof on my part. And, since I figured I was going to finish the proof anyway, I decided I would write about it in my writeup of assignment 8 as well.


Our story begins with a simple acute triangle. We construct each of the altitudes for the sides of the triangle, and label their intersection point the orthocenter.


Our proof is about this equation and its result:

I used GSP to explore this relationship and found that it holds for a variety of acute triangles. But, as we all know, a GSP demonstration is not a proof, so I guess we should write one.


First, it would be a good idea to make some triangles and to draw the circumcircle of our triangle. We will also label the points of intersection from our altitudes and the circle.

Lets begin our proof by building some relationships between triangles. The triangles which are colored alike can be shown to be similar. Each pair of triangles contains a right angle from the altitude, and a second angle pair will be congruent by vertical angles. This means triangles AFH and CDH, triangles AEH and BDH, and triangles HFB and HEC are all similar by AA similarity. We can use the definition of similarity to conclude that all three angles of all pairs of triangles are congruent.

This tells us something about angles, but nothing about sides, so we are going to draw some construction lines and look for more triangle relationships.

We can show that each pair of numbered triangles are congruent by ASA. Let me break it down for you...

Let's begin with triangles AHE and AQE. We know that they share segment AE, and that they both contain right angles at E. To show that angles HAE and QAE are congruent, we need to be a little sneaky.

As you can see in the diagram, the angles labeled as "a" will be congruent. We have already shown that HAE is congruent to HBD from the similar triangles. Now we will use the fact that QBC and QAC both subtend the same arc of the circle. Since this is true, those angles must be congruent. By transitivity, we can state that HAE and QAE are congruent, thus giving us the triangles congruent by ASA.


Using similar arguments, we can show the other pairs of triangles congruent as well. In these cases, the angles marked as "b" will all be congruent by the same combination of similarity of triangles, inscribed angles subtending the same arc, and the transitive property.

So, what is the point of all of this? It is all prelude to the main event, but necessary so we can state one crucial fact. That is that since the pairs of triangles are congruent, the definition of congruency tells us that the corresponding parts are all congruent. In this case, what we need is that:

HE = EQ, HD = DP, and HF = FR

Now we can get back to our original goal, proving that

We begin by using segment addition to rewrite some of our values, namely:

AP = AD + DP

QB = BE + EQ

RC = RF + FC


Now we use substitution from our above equalities to write:

AP = AD + DP

QB = BE + EH

RC = CF + FH

Which means our new fractions look like this:

Which means our original equation becomes

We want this to equal 4, so we can infer that we need to prove that

In order to do this we are going to need to approach it a bit differently than just manipulating fractions. We will use what we know about area of these triangles to prove this statement. We begin by recoloring our triangles to show that the area of the three small triangles will equal the area of triangle ABC.

We know that the area of a triangle equals 1/2 bh. And we know that since areas of polygons are additive, we can write the following equations:


If we multiply both sides by 2 to eliminate fractions, we see that

We want this to equal 1, so we divide both sides by 2 times the area of ABC

Now, we know that the area of triangle ABC can be calculated by using 1/2 bh, so there are three different ways we can express the area, and more specifically, two times the area.

So, lets use substitution into the above fractions, using the appropriate value of the area of triangle ABC to ensure our fractions become what we desire.

You should notice that in each fraction, we will be able to reduce by a common factor, this leaves us with:

Which, is exactly what we needed to show. Therefore, we can conclude that for all acute triangles:


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